Question

A variable straight line through the point
of intersection of the two lines x/3 + y/2 = 1
and x/2 +y/3= 1 meets the co-ordinate
axes at A and B. find the locus of
the middle point of AB.​

Answer 1

Answer:

a point of ab 3=1y2+ab answer

Answer 2

Given straight lines.

$\longrightarrow \dfrac{x}{3}+\dfrac{y}{2}=1$

$\longrightarrow2x+3y-6=0$

And,

$\longrightarrow \dfrac{x}{2}+\dfrac{y}{3}=1$

$\longrightarrow3x+2y-6=0$

Let the equation of the variable straight line through the point of intersection of the two given lines be,

$\longrightarrow(2x+3y-6)+\lambda(3x+2y-6)=0$

$\longrightarrow2x+3y-6+3\lambda\,x+2\lambda\,y-6\lambda=0$

$\longrightarrow(2+3\lambda)x+(3+2\lambda)y-6(1+\lambda)=0$

$\longrightarrow(2+3\lambda)x+(3+2\lambda)y=6(1+\lambda)$

$\longrightarrow\dfrac{(2+3\lambda)x}{6(1+\lambda)}+\dfrac{(3+2\lambda)y}{6(1+\lambda)}=1$

$\longrightarrow\dfrac{x}{\left(\dfrac{6(1+\lambda)}{2+3\lambda}\right)}+\dfrac{y}{\left(\dfrac{6(1+\lambda)}{3+2\lambda}\right)}=1$

Now the equation is in intercept form, from which we can assume that,

$\longrightarrow A=\left(\dfrac{6(1+\lambda)}{2+3\lambda},\ 0\right)$

$\longrightarrow B=\left(0,\ \dfrac{6(1+\lambda)}{3+2\lambda}\right)$

since A and B are intercepts.

Then midpoint of AB is,

$\longrightarrow C=\left(\dfrac{\dfrac{6(1+\lambda)}{2+3\lambda}+0}{2},\ \dfrac{0+\dfrac{6(1+\lambda)}{3+2\lambda}}{2}\right)$

$\longrightarrow C=\left(\dfrac{3(1+\lambda)}{2+3\lambda},\ \dfrac{3(1+\lambda)}{3+2\lambda}\right)$

Let us find the locus of this point by taking $C=(x,\ y),$ then we have,

$\longrightarrow x=\dfrac{3(1+\lambda)}{2+3\lambda}$

$\longrightarrow 2x+3\lambda\ x=3+3\lambda$

$\longrightarrow3\lambda\ x-3\lambda=3-2x$

$\longrightarrow\lambda=\dfrac{3-2x}{3x-3}\quad\quad\dots(1)$

And,

$\longrightarrow y=\dfrac{3(1+\lambda)}{3+2\lambda}$

From (1),

$\longrightarrow y=\dfrac{3\left(1+\dfrac{3-2x}{3x-3}\right)}{3+2\left(\dfrac{3-2x}{3x-3}\right)}$

$\longrightarrow y=\dfrac{\left(\dfrac{3x-3+3-2x}{x-1}\right)}{\dfrac{1}{3}\left(\dfrac{9x-9+6-4x}{x-1}\right)}$

$\longrightarrow y=\dfrac{3x}{5x-3}$

$\longrightarrow\underline{\underline{5xy-3y-3x=0}}$